alt.energy.renewable

wrote:

>> You might make a U-shaped strawbale platform with the entrance door into
>> the opening of the U, inside an icosahedron made with 15 ' equilateral
>> triangles cut from 5 4'x8' sheets of foil foamboard with some foam in a can
>> for glue at the seams and white paint over it all. That would have
>> a /tan(36) = ' inner diameter and a 4' stemwall height and a peak
>> 0.526x = ' above that. Pretty big.
>>
>> You could make a smaller version with more cutting and piecing. Maybe
>> divide each large triangle into 4 smaller triangles, conceptually, and
>> divide the foamboard into thirds lengthwise to make 5 big triangles
>> and 1 3/4 triangle and 4 small triangles and 6 half-small triangles,
>> like this, cut from 2 4x8 sheets, viewed in a fixed font:
>>
>> ----------------------------- The big triangles would have '
>> |1/ \ / \ / \2| edges, and the dome would have
>> |/ 1 \ / \ / 2 \| a ' ID and a ' stemwall
>> |-----\ 1 / 2 \ 3 /-----| height and a peak ' above.
>> |\ 3 / \ / \ / \ 4 /|
>> |3\ / \ / \ / \ /4|
>> |--/ \-----------/ \--|
>> |5/ 4 \ 1 / 5 \6|
>> |/ \ / \|
>> -----------------------------
>
>I'll think about that.

Oops. The big triangles would be 3/cos(30) = ' on a side, and the dome
would have a ' ID and a 3' stemwall and a peak ' above that, with
64 ft^2 of surface, vs 80 ft^2 for a 4' cube, not counting the floor.

>The thing I'm for the next few days is to insulate over the south
>window on the inside with R12 and see how this does with just
>simulated dog heat...

A 4' R12 cube with G = 6x4'x4'/R12 = 8 Btu/h-F and a 40 watt bulb and
no air leaks might be /8 = F warmer than the outdoors...
5 cfm of fresh air would make it roughly /13 = F warmer.

With indirect gain, if 1020 Btu/ft^2 falls on a south wall on an average
January day with a max in Billings, a 4' R12 cube with 16 ft^2
of R2 sunspace glazing with 80% solar transmission over an air gap over
an R12 insulated south wall with lots of shiny ceiling mass and surface
and a 65 F average living space temp would have +x24h
= 16332 Btu = 6h()16/2 [for the south wall during the day]
+18h(65-)16/18 [for the south wall at night] +24h(T-)16/12 [for
the ceiling] +24h(65-)4x16/12 [for the other 3 walls and the floor]
Btu/day, with ceiling mass temp T = 154 F, if I did that right.

If the average ceiling mass temp is (154+65)/2 = over 5 cloudy days
while the cube loses 5x24((65-)5x16/12+(-)16/12-)
= 31254 Btu = (154-65)C, ceiling mass C = 351 Btu/F, eg a " x 16 ft^2
layer of water. Another layer of R12 ceiling insulation would help.

Nick